\problemdata      
        {summations-5} % latex-friendly label for the prob.
        {Summations} % The topic(s) of the problem content.
        {?} % Source (if known)
        {S02 PS6} % Usage (list ones you are aware of).
        {Christos Kapoutsis, S02} % Last revision info (author, date).


\begin{problem}
Use the integral method to find upper and lower bounds that differ by at
most 0.1 for the following sum
\iffalse
Note that you may need to add the first
few terms explicitly and then use integrals to bound the sum of the
remaining terms.
\fi
\[
\sum_{i=1}^{\infty} \frac{1}{i^2}.
\]

\solution{
We can bound the summation above as follows:
\begin{eqnarray*}
\sum_{i=1}^{\infty} \frac{1}{i^2}
	& \leq &	\frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
			\int_3^{\infty} \frac{1}{x^2} \ dx \\
	& = &		\frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
			\left( - \frac{1}{x} \right)_3^{\infty} \\
	& = &		\frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
			\frac{1}{3} \\
	& = &		1.694\ldots
\end{eqnarray*}
We can bound the summation below similarly:
\begin{eqnarray*}
\sum_{i=1}^{\infty} \frac{1}{i^2}
	& \geq &	\frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
			\int_3^{\infty} \frac{1}{(x+1)^2} \ dx \\
	& = &		\frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
			\left( - \frac{1}{x+1} \right)_3^{\infty} \\
	& = &		\frac{1}{1} + \frac{1}{4} + \frac{1}{9} +
			\frac{1}{4} \\
	& = &		1.611\ldots
\end{eqnarray*}
}
\noindent
(The actual value of the summation turns out to be $\pi^2 / 6 =
1.644\ldots$.)
\end{problem}

